# How much work would it take to horizontally accelerate an object with a mass of 8 kg to 4 ms^-1 on a surface with a kinetic friction coefficient of 1 ?

Apr 16, 2016

Not enough information provided.

#### Explanation:

We can divide the total work into two parts,
1. To increase the kinetic energy of the object. (Which is known with the formula $\frac{1}{2} m {v}^{2}$)

1. The work done against the frictional force. (Given as $\mu \cdot m g \cdot s$; unknown as no other parameter except $v$ (final velocity) is provided).
Apr 16, 2016

$\left(64 - 156.8 t\right) N$, where $t$ is the time of acceleration.
With the condition that $t > 0.4082 s$

#### Explanation:

Force $F$ applied to accelerate the object$= m . a$
Assuming that the object is initially at rest on the horizontal and acceleration is constant. Using the kinematic equation

$v = u + a t$
$4 = a t$
$a = \frac{4}{t}$
Also distance moved $s = u t + \frac{1}{2} a {t}^{2}$
$s = \frac{1}{2} \times \frac{4}{t} \times {t}^{2} = 2 t$

Given is coefficient of kinetic friction $\mu = 1$.
As normally coefficient of static friction is greater than the coefficient of kinetic friction, also that neither of the coefficients can be greater than $1$, implies that both the coefficients ${\mu}_{s} \mathmr{and} {\mu}_{k} = 1$.

Let the normal reaction be $N = m g$, where gravity $g = 9.8 m {s}^{-} 2$.
Force of friction which will oppose the motion ${F}_{f} = \mu . m g$
Inserting given values
${F}_{f} = 1 \times 8 \times 9.8 = 78.4 N$
$\text{Net Force"="Applied Force"-"Force due to friction}$
$\text{Net Force} = 8 \times \frac{4}{t} - 78.4$
$= \frac{32}{t} - 78.4$

Work done $= \text{Force"xx"Distance}$

$= \left(\frac{32}{t} - 78.4\right) \times 2 t$
$= 64 - 156.8 t$